1)What is the magnitude of the electric field at point P, located at (6.50 cm, 0

Question

1)What is the magnitude of the electric field at point P, located at (6.50 cm, 0), due to Q1 alone?
3.71×106 N/C

2)What is the x-component of the total electric field at P?
1.51×107 N/C

3) What is the y-component of the total electric field at P?

4) What is the magnitude of the total electric field at P?

6) Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Explanation / Answer

1) Required E(magnitude) = [(9*10^9)*(2.00*10^-6)]/ [(2.5^2 +6.50^2)*10^-4]
E = (18*10^3)/(48.5*10^-4) = 3.7*10^6 N/C
E will make an angle of tan^-1[2.5/6.5] = 21 degree with positive x-axis

2) X component, TEx of Total electric field

                      = [{(9*10^9)*(6.7*10^-6)}/[{(2.5^2 +6.5^2)*10^-4}] cos21 + (3.7*10^6)*cos 21
TEx = [(60.3*10^3)/(48.5*10^-4) + (3.7*10^6)]*cos21 = [(12.4+3.7)*cos21]*10^6 N = 15.03*10^6 N/C
TEx will be along positive x axis

3)Similarly TEy = {12.4 - 3.7]*sin21*10^6 = 3.12*10^6 N/C
Direction of TEy will be along negative y direction

4)Total electric field, TE = (10^6)*sqrt[15.03^2 + 3.12^2] = 15.35*10^6 N/C
Direction of TE will make an angle of tan^-1[3.12/15.03] = 11.7 degree with positive x-axis in clockwise direction. vector lies in 4th quadrant, where x-component is positive and y-component is negative

5)TE(symmetric) = 2*(3.7*10^6)*cos21 = 6.9*10^6 N/C
[y components cancel because of symmetry and contribution of x components are equal from both charges]

6)Required force = (6.9*10^6)*(1.6*10^-19) = 1.104*10^-12 N along x-axis towards origin.

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